19.5.1 Temperature Change and Specific Heat

Suppose you do an experiment in which you add energy to water, either by doing work on it or by transferring heat to it. Either way, you will find that adding $4190 J$ of energy raises the temperature of 1 kg of water by 1 K. If you were fortunate enough to have 1 kg of gold, you would need to add only 129 J of energy to raise its temperature by 1 K.

The amount of energy that raises the temperature of 1 kg of a substance by 1 K is called the specific heat of that substance. The symbol for specific heat is c. Water has specific heat $c_{water} = 4190 J / kg K$ . The specific heat of gold is $c_{gold} = 129 J / kg K$ . Specific heat depends only on the material from which an object is made. TABLE 19.2 provides some specific heats for common liquids and solids.

TABLE 19.2 Specific heats and molar specific heats of solids and liquids

Substance	$c$ (J/kg K)	$C$ (J/mol K)
Solids
Aluminum	900	24.3
Copper	385	24.4
Iron	449	25.1
Gold	129	25.4
Lead	128	26.5
Ice	2090	37.6
Liquids
Ethyl alcohol	2400	110.4
Mercury	140	28.1
Water	4190	75.4

NOTE The term “specific heat” does not use the word “heat” in the way that we have defined it. Specific heat is an old idea, dating back to the days of the caloric theory when heat was thought to be a substance contained in the object. The term has continued in use even though our understanding of heat has changed.

If energy $c$ is required to raise the temperature of 1 kg of a substance by 1 K, then energy Mc is needed to raise the temperature of mass $M$ by 1 K and $(M c) Δ T$ is needed to raise the temperature of mass $M$ by $Δ T$ . In other words, the thermal energy of the system changes by

Δ E_{th} = M c Δ T (temperature change)

(19.15)

when its temperature changes by $Δ T$ . $Δ E_{th}$ can be either positive (thermal energy increases as the temperature goes up) or negative (thermal energy decreases as the temperature goes down). Recall that uppercase $M$ is used for the mass of an entire system while lowercase $m$ is reserved for the mass of an atom or molecule.

NOTE In practice, $Δ T$ is usually measured in $° C$ . But the Kelvin and the Celsius temperature scales have the same step size, so $Δ T$ in K has exactly the same numerical value as $Δ T$ in $° C$ . Thus

You do not need to convert temperatures from $° C$ to K if you need only a temperature change $Δ T$ .
You do need to convert anytime you need the actual temperature T.

The first law of thermodynamics, $Δ E_{th} = W + Q$ , allows us to write Equation 19.15 as $M c Δ T = W + Q$ . In other words, we can change the system’s temperature either by heating it or by doing an equivalent amount of work on it. In working with solids and liquids, we almost always change the temperature by heating. If $W = 0$ , which we will assume for the rest of this section, then the heat energy needed to bring about a temperature change $Δ T$ is

Q = M c Δ T (temperature change)

(19.16)

Because $Δ T = Δ E_{th} / M c$ , it takes more energy to change the temperature of a substance with a large specific heat than to change the temperature of a substance with a small specific heat. You can think of specific heat as measuring the thermal inertia of a substance. Metals, with small specific heats, warm up and cool down quickly. A piece of aluminum foil can be safely held within seconds of removing it from a hot oven. Water, with a very large specific heat, is slow to warm up and slow to cool down. This is fortunate for us. The large thermal inertia of water is essential for the biological processes of life. We wouldn’t be here studying physics if water had a small specific heat!

EXAMPLE 19.3 Running a fever

A $70 kg$ student catches the flu, and his body temperature increases from $37.0 ° C (98.6 ° F) to 39.0 ° C (102.2 ° F)$ . How much energy is required to raise his body’s temperature? The specific heat of a mammalian body is $3400 J / kg K$ , nearly that of water because mammals are mostly water.

MODEL Energy is supplied by the chemical reactions of the body’s metabolism. These exothermic reactions transfer heat to the body. Normal metabolism provides enough heat energy to offset energy losses (radiation, evaporation, etc.) while maintaining a normal body temperature of $37 ° C$ . We need to calculate the additional energy needed to raise the body’s temperature by $2.0 ° C$ , or 2.0 K.

SOLVE The necessary heat energy is

Q = M c Δ T = (70 kg) (3400 J/kg K) (2.0 K) = 4.8 \times 10^{5} J

REVIEW This appears to be a lot of energy, but a joule is actually a very small amount of energy. It is only 110 Cal, approximately the energy gained by eating an apple.

The molar specific heat is the amount of energy that raises the temperature of 1 mol of a substance by 1 K. We’ll use an uppercase $C$ for the molar specific heat. The heat energy needed to bring about a temperature change $Δ T$ of $n$ moles of substance is

Q = n C Δ T

(19.17)

Molar specific heats are listed in Table 19.2. Look at the five elemental solids (excluding ice). All have $C$ very near $25 J / mol K$ . If we were to expand the table, we would find that most elemental solids have $C \approx 25 J / mol K$ . This can’t be a coincidence, but what is it telling us? This is a puzzle we will address in Chapter 20, where we will explore thermal energy at the atomic level.